YES
We show the termination of the TRS R:

  app(nil(),k) -> k
  app(l,nil()) -> l
  app(cons(x,l),k) -> cons(x,app(l,k))
  sum(cons(x,nil())) -> cons(x,nil())
  sum(cons(x,cons(y,l))) -> sum(cons(plus(x,y),l))
  sum(app(l,cons(x,cons(y,k)))) -> sum(app(l,sum(cons(x,cons(y,k)))))
  plus(|0|(),y) -> y
  plus(s(x),y) -> s(plus(x,y))


-- SCC decomposition.

Consider the dependency pair problem (P, R), where P consists of

p1: app#(cons(x,l),k) -> app#(l,k)
p2: sum#(cons(x,cons(y,l))) -> sum#(cons(plus(x,y),l))
p3: sum#(cons(x,cons(y,l))) -> plus#(x,y)
p4: sum#(app(l,cons(x,cons(y,k)))) -> sum#(app(l,sum(cons(x,cons(y,k)))))
p5: sum#(app(l,cons(x,cons(y,k)))) -> app#(l,sum(cons(x,cons(y,k))))
p6: sum#(app(l,cons(x,cons(y,k)))) -> sum#(cons(x,cons(y,k)))
p7: plus#(s(x),y) -> plus#(x,y)

and R consists of:

r1: app(nil(),k) -> k
r2: app(l,nil()) -> l
r3: app(cons(x,l),k) -> cons(x,app(l,k))
r4: sum(cons(x,nil())) -> cons(x,nil())
r5: sum(cons(x,cons(y,l))) -> sum(cons(plus(x,y),l))
r6: sum(app(l,cons(x,cons(y,k)))) -> sum(app(l,sum(cons(x,cons(y,k)))))
r7: plus(|0|(),y) -> y
r8: plus(s(x),y) -> s(plus(x,y))

The estimated dependency graph contains the following SCCs:

  {p4}
  {p1}
  {p2}
  {p7}


-- Reduction pair.

Consider the dependency pair problem (P, R), where P consists of

p1: sum#(app(l,cons(x,cons(y,k)))) -> sum#(app(l,sum(cons(x,cons(y,k)))))

and R consists of:

r1: app(nil(),k) -> k
r2: app(l,nil()) -> l
r3: app(cons(x,l),k) -> cons(x,app(l,k))
r4: sum(cons(x,nil())) -> cons(x,nil())
r5: sum(cons(x,cons(y,l))) -> sum(cons(plus(x,y),l))
r6: sum(app(l,cons(x,cons(y,k)))) -> sum(app(l,sum(cons(x,cons(y,k)))))
r7: plus(|0|(),y) -> y
r8: plus(s(x),y) -> s(plus(x,y))

The set of usable rules consists of

  r1, r2, r3, r4, r5, r6, r7, r8

Take the reduction pair:

  generalized weighted path order (>=MG, >G)
  
    base order 1:
  
      trivial reduction triple (T x T, T x T, Φ)
  
    base order 2:
  
      lexicographic combination of reduction triples:
      
        1. marked-algebra reduction triple (>=A, >=A!, >A!):
        
          max/plus interpretation on natural numbers:
          
            sum#_A(x1) = max{0, x1 - 24}
            app_A(x1,x2) = max{23, x1 + 17, x2 + 1}
            cons_A(x1,x2) = max{11, x2}
            sum_A(x1) = 22
            nil_A = 4
            plus_A(x1,x2) = max{39, x1 - 35, x2 + 37}
            |0|_A = 0
            s_A(x1) = max{0, x1 - 40}
            sum#!_A(x1) = x1 + 16
            app!_A(x1,x2) = max{x1 + 40, x2 + 7}
            cons!_A(x1,x2) = max{8, x2 + 3}
            sum!_A(x1) = max{2, x1 - 2}
            nil!_A = 1
            plus!_A(x1,x2) = max{15, x1 - 2, x2 - 2}
            |0|!_A = 0
            s!_A(x1) = max{1, x1 - 40}
        
        2. marked-algebra reduction triple (>=A, >=A!, >A!):
        
          max/plus interpretation on natural numbers:
          
            sum#_A(x1) = max{0, x1 - 21}
            app_A(x1,x2) = max{x1 + 1, x2 + 17}
            cons_A(x1,x2) = max{16, x1 - 18, x2}
            sum_A(x1) = max{5, x1}
            nil_A = 6
            plus_A(x1,x2) = max{24, x1 - 4, x2}
            |0|_A = 0
            s_A(x1) = max{1, x1 - 25}
            sum#!_A(x1) = x1 + 4
            app!_A(x1,x2) = max{x1 + 43, x2 + 25}
            cons!_A(x1,x2) = max{0, x2 - 15}
            sum!_A(x1) = max{26, x1 - 7}
            nil!_A = 10
            plus!_A(x1,x2) = max{10, x1 - 23, x2 - 24}
            |0|!_A = 0
            s!_A(x1) = max{1, x1 - 45}
        
  
    partial status:
  
      pi(sum#) = [1]
      pi(app) = [1, 2]
      pi(cons) = [2]
      pi(sum) = []
      pi(nil) = []
      pi(plus) = [2]
      pi(|0|) = []
      pi(s) = []

The following rules are removed:

  p1

Then no dependency pair remains.

-- Reduction pair.

Consider the dependency pair problem (P, R), where P consists of

p1: app#(cons(x,l),k) -> app#(l,k)

and R consists of:

r1: app(nil(),k) -> k
r2: app(l,nil()) -> l
r3: app(cons(x,l),k) -> cons(x,app(l,k))
r4: sum(cons(x,nil())) -> cons(x,nil())
r5: sum(cons(x,cons(y,l))) -> sum(cons(plus(x,y),l))
r6: sum(app(l,cons(x,cons(y,k)))) -> sum(app(l,sum(cons(x,cons(y,k)))))
r7: plus(|0|(),y) -> y
r8: plus(s(x),y) -> s(plus(x,y))

The set of usable rules consists of

  (no rules)

Take the reduction pair:

  generalized weighted path order (>=MG, >G)
  
    base order 1:
  
      trivial reduction triple (T x T, T x T, Φ)
  
    base order 2:
  
      lexicographic combination of reduction triples:
      
        1. marked-algebra reduction triple (>=A, >=A!, >A!):
        
          max/plus interpretation on natural numbers:
          
            app#_A(x1,x2) = max{0, x1 - 6}
            cons_A(x1,x2) = max{x1 - 53, x2 + 5}
            app#!_A(x1,x2) = max{1, x1 - 48, x2 - 1}
            cons!_A(x1,x2) = max{3, x1 - 140, x2 - 47}
        
        2. marked-algebra reduction triple (>=A, >=A!, >A!):
        
          max/plus interpretation on natural numbers:
          
            app#_A(x1,x2) = max{0, x1 - 4}
            cons_A(x1,x2) = max{x1 - 49, x2 + 3}
            app#!_A(x1,x2) = max{4, x1 - 4, x2 - 1}
            cons!_A(x1,x2) = max{3, x1 - 126, x2 - 133}
        
  
    partial status:
  
      pi(app#) = [1]
      pi(cons) = [2]

The following rules are removed:

  p1

Then no dependency pair remains.

-- Reduction pair.

Consider the dependency pair problem (P, R), where P consists of

p1: sum#(cons(x,cons(y,l))) -> sum#(cons(plus(x,y),l))

and R consists of:

r1: app(nil(),k) -> k
r2: app(l,nil()) -> l
r3: app(cons(x,l),k) -> cons(x,app(l,k))
r4: sum(cons(x,nil())) -> cons(x,nil())
r5: sum(cons(x,cons(y,l))) -> sum(cons(plus(x,y),l))
r6: sum(app(l,cons(x,cons(y,k)))) -> sum(app(l,sum(cons(x,cons(y,k)))))
r7: plus(|0|(),y) -> y
r8: plus(s(x),y) -> s(plus(x,y))

The set of usable rules consists of

  r7, r8

Take the reduction pair:

  generalized weighted path order (>=MG, >G)
  
    base order 1:
  
      trivial reduction triple (T x T, T x T, Φ)
  
    base order 2:
  
      lexicographic combination of reduction triples:
      
        1. marked-algebra reduction triple (>=A, >=A!, >A!):
        
          max/plus interpretation on natural numbers:
          
            sum#_A(x1) = max{0, x1 - 17}
            cons_A(x1,x2) = max{x1 - 2, x2 + 7}
            plus_A(x1,x2) = max{18, x1 - 9, x2 + 1}
            |0|_A = 0
            s_A(x1) = max{2, x1 - 19}
            sum#!_A(x1) = max{1, x1 - 45}
            cons!_A(x1,x2) = max{25, x1 - 52, x2 - 42}
            plus!_A(x1,x2) = max{26, x1 - 3, x2 - 1}
            |0|!_A = 0
            s!_A(x1) = max{1, x1 - 41}
        
        2. marked-algebra reduction triple (>=A, >=A!, >A!):
        
          max/plus interpretation on natural numbers:
          
            sum#_A(x1) = max{0, x1 - 19}
            cons_A(x1,x2) = max{x1 - 27, x2 + 5}
            plus_A(x1,x2) = max{45, x1 - 47, x2 + 1}
            |0|_A = 0
            s_A(x1) = max{2, x1 - 46}
            sum#!_A(x1) = max{1, x1 - 62}
            cons!_A(x1,x2) = max{3, x1 - 135, x2 - 60}
            plus!_A(x1,x2) = max{4, x1 - 3, x2 - 40}
            |0|!_A = 0
            s!_A(x1) = max{1, x1 - 95}
        
  
    partial status:
  
      pi(sum#) = [1]
      pi(cons) = [2]
      pi(plus) = [2]
      pi(|0|) = []
      pi(s) = []

The following rules are removed:

  p1

Then no dependency pair remains.

-- Reduction pair.

Consider the dependency pair problem (P, R), where P consists of

p1: plus#(s(x),y) -> plus#(x,y)

and R consists of:

r1: app(nil(),k) -> k
r2: app(l,nil()) -> l
r3: app(cons(x,l),k) -> cons(x,app(l,k))
r4: sum(cons(x,nil())) -> cons(x,nil())
r5: sum(cons(x,cons(y,l))) -> sum(cons(plus(x,y),l))
r6: sum(app(l,cons(x,cons(y,k)))) -> sum(app(l,sum(cons(x,cons(y,k)))))
r7: plus(|0|(),y) -> y
r8: plus(s(x),y) -> s(plus(x,y))

The set of usable rules consists of

  (no rules)

Take the reduction pair:

  generalized weighted path order (>=MG, >G)
  
    base order 1:
  
      trivial reduction triple (T x T, T x T, Φ)
  
    base order 2:
  
      lexicographic combination of reduction triples:
      
        1. marked-algebra reduction triple (>=A, >=A!, >A!):
        
          max/plus interpretation on natural numbers:
          
            plus#_A(x1,x2) = max{0, x1 - 2}
            s_A(x1) = max{52, x1 + 1}
            plus#!_A(x1,x2) = max{2, x1 - 125, x2 - 1}
            s!_A(x1) = max{1, x1 - 124}
        
        2. marked-algebra reduction triple (>=A, >=A!, >A!):
        
          max/plus interpretation on natural numbers:
          
            plus#_A(x1,x2) = max{0, x1 - 2}
            s_A(x1) = max{33, x1 + 1}
            plus#!_A(x1,x2) = max{7, x1 - 36, x2 - 1}
            s!_A(x1) = max{6, x1 - 35}
        
  
    partial status:
  
      pi(plus#) = [1]
      pi(s) = [1]

The following rules are removed:

  p1

Then no dependency pair remains.